Differentiation takes a function and gives its slope. Integration runs that backwards: given the slope everywhere, rebuild the function. The same idea measures area. Add up thin strips under a curve and the total turns out to be the rise of the rebuilt function. One idea, two uses: indefinite integrals (families of functions) and definite integrals (numbers: areas, distances, totals).
Differentiation turns a function into its slope. Integration goes the other way: from the slope, rebuild the function.
Read it as: “which function has derivative f(x)?” The answer F is called an antiderivative (or primitive) of f, so it works whenever F′(x) = f(x). The f inside is the integrand, and x is the variable of integration.
A constant disappears when you differentiate, so any constant C can be added: the constant of integration. Two antiderivatives of f on an interval differ only by a constant.
So integrate term by term and keep constants outside. Also d/dx ∫ f dx = f.
Powers and logs
Exponentials
Trigonometric
Inverse trigonometric
Each tile is a derivative read backwards: the standard integrals. Add + C to every answer.
∫xⁿ dx = xⁿ⁺¹/(n + 1) fails at n = −1 (division by 0); ∫x⁻¹ dx is log|x|. And some functions, such as e^{−x²}, have antiderivatives that cannot be written with the usual functions at all.
Antiderivatives by inspection of cos 2x, 3x² + 4x³, 1/x.
(x³ − 1)/x², x^{2/3} + 1, x^{3/2} + 2eˣ − 1/x.
sin x + cos x; cosec x(cosec x + cot x); (1 − sin x)/cos²x.
F′ = 4x³ − 6 with F(0) = 3.
If the integrand contains a function and its derivative, give that function a new name, t. The integral becomes a simpler one in t.
You can also put t = g(x). In practice: look for an inside function whose derivative also appears as a factor, and call it t.
First use an identity to lower the power or split the product (Section 7.3.2), then integrate.
| f(x) | ∫ f(x) dx |
|---|---|
| tan x | log |sec x| |
| cot x | log |sin x| |
| sec x | log |sec x + tan x| |
| cosec x | log |cosec x − cot x| |
sin mx; 2x sin(x² + 1); tan⁴√x sec²√x/√x; sin(tan⁻¹x)/(1 + x²).
sin³x cos²x; sin x/sin(x + a); 1/(1 + tan x).
cos²x; sin 2x cos 3x; sin³x.
Six integrals come up again and again. Learn them once; then most quadratic denominators can be turned into one of them.
| f(x) | ∫ f(x) dx | |
|---|---|---|
| 1 | 1x² − a² | 12a log |x − a||x + a| |
| 2 | 1a² − x² | 12a log |a + x||a − x| |
| 3 | 1x² + a² | 1a tan⁻¹ xa |
| 4 | 1√(x² − a²) | log |x + √(x² − a²)| |
| 5 | 1√(a² − x²) | sin⁻¹ xa |
| 6 | 1√(x² + a²) | log |x + √(x² + a²)| |
How to use them
1/(x² − 16); 1/√(2x − x²).
1/(x² − 6x + 13); 1/(3x² + 13x − 10); 1/√(5x² − 2x).
(x + 2)/(2x² + 6x + 5); (x + 3)/√(5 − 4x − x²).
A hard fraction can often be split into a sum of simple fractions. Each simple piece is easy to integrate.
A fraction of two polynomials is a rational function. It is proper when the degree of P is less than the degree of Q.
Long division gives a polynomial T(x) plus a proper fraction. Then split only the proper part.
| Fraction, then how to split it | |
|---|---|
| 1 | px + q(x − a)(x − b)= Ax − a + Bx − b |
| 2 | px + q(x − a)²= Ax − a + B(x − a)² |
| 3 | px² + qx + r(x − a)(x − b)(x − c)= Ax − a + Bx − b + Cx − c |
| 4 | px² + qx + r(x − a)²(x − b)= Ax − a + B(x − a)² + Cx − b |
| 5 | px² + qx + r(x − a)(x² + bx + c)= Ax − a + Bx + Cx² + bx + c |
How to use it
1/((x + 1)(x + 2)).
(x² + 1)/(x² − 5x + 6).
(3x − 2)/((x + 1)²(x + 3)).
x²/((x² + 1)(x² + 4)).
(3 sin φ − 2) cos φ/(5 − cos²φ − 4 sin φ).
(x² + x + 1)/((x + 2)(x² + 1)).
Use it when the integrand is a product of two functions. It swaps the hard integral for an easier one.
Here f = f(x), g = g(x). In words: first × integral of the second, minus the integral of (derivative of the first × integral of the second).
Choose as first the function that gets simpler when differentiated. A useful order is ILATE: Inverse trig, Log, Algebraic, Trig, Exponential.
A lone log x or tan⁻¹ x is a product with the function 1. Take 1 as the second function.
When eˣ multiplies a function plus its own derivative, write the answer down at once.
| f(x) | ∫ f(x) dx |
|---|---|
| √(x² − a²) | x2 √(x² − a²) − a²2 log |x + √(x² − a²)| |
| √(x² + a²) | x2 √(x² + a²) + a²2 log |x + √(x² + a²)| |
| √(a² − x²) | x2 √(a² − x²) + a²2 sin⁻¹ xa |
x cos x.
log x.
xeˣ.
x sin⁻¹x/√(1 − x²).
eˣ sin x.
eˣ(tan⁻¹x + 1/(1 + x²)); (x² + 1)eˣ/(x + 1)².
√(x² + 2x + 5).
√(3 − 2x − x²).
A definite integral is a number: the area under a curve between two lines. Any antiderivative gives it with one subtraction.
The area between y = f(x), the x-axis and the lines x = a and x = b. Here a is the lower limit and b the upper limit. Parts below the axis count as negative. It is defined as the limit of sums of thin strips.
A(x) is the area from a up to x. First fundamental theorem: its slope at x is the height f(x).
Second fundamental theorem. F is any antiderivative of f, and f must be defined and continuous on the whole of [a, b]. Written [F(x)]ₐᵇ. The constant C cancels, so leave it out.
Substitution in definite integrals
∫₂³ x²; ∫₄⁹ √x/(30 − x^{3/2})²; ∫₁² x/((x + 1)(x + 2)); ∫₀^{π/4} sin³2t cos 2t.
∫₋₁¹ 5x⁴√(x⁵ + 1) dx.
∫₀¹ tan⁻¹x/(1 + x²) dx.
Eight properties, P₀ to P₇, let you rearrange a definite integral before you work it out. Many hard integrals become easy.
P₀ the letter does not matter. P₁ swapping the limits changes the sign. P₂ split the interval at a point c.
P₃ and P₄: replace x by its mirror image in the interval and the integral stays the same.
P₅, then P₆: it is 2 ∫₀ᵃ f if f(2a − x) = f(x), and 0 if f(2a − x) = − f(x).
P₇: it is 2 ∫₀ᵃ f if f is even, and 0 if f is odd.
The mirror-and-add trick (P₃, P₄)
∫₋₁² |x³ − x| dx.
∫ sin²x from −π/4 to π/4.
∫₀^π x sin x/(1 + cos²x) dx.
∫₋₁¹ sin⁵x cos⁴x dx.
∫₀^{π/2} sin⁴x/(sin⁴x + cos⁴x) dx.
∫ dx/(1 + √tan x) from π/6 to π/3.
∫₀^{π/2} log sin x dx.
cos 6x √(1 + sin 6x).
(x⁴ − x)^{1/4}/x⁵.
x⁴/((x − 1)(x² + 1)).
log(log x) + 1/(log x)².
√cot x + √tan x.
sin 2x cos 2x/√(9 − cos⁴2x).
∫₋₁^{3/2} |x sin πx| dx.
∫₀^π x/(a²cos²x + b²sin²x) dx.
Every MCQ from Exercises 7.1–7.10 and the miscellaneous exercise, plus two concept checks. Tap an option to see the reasoning.
when F′ = f. Antiderivatives of the same function differ by a constant.
Every derivative formula gives a standard integral.
with t = g(x). Identities lower powers and split products.
Complete the square to reach the six standard forms; split px + q into A (derivative of the quadratic) + B.
Partial fractions for rational functions (divide first if improper). By parts for products.
is the signed area, and A′(x) = f(x). Symmetry properties P₀–P₇ shorten many definite integrals.
Found the area of a parabolic segment by filling it with ever smaller triangles and adding them up: the first "limit of a sum".
Cut regions into thin slices ("indivisibles") and found ∫xⁿ dx for many n by adding the slices.
Saw that finding areas undoes finding tangents: the Fundamental Theorem. Leibniz introduced the ∫ sign, a long S for "summa", and the dx.
Gave the modern definition of the definite integral as a limit of sums of strips, the one used in this chapter.
Each example is a small animated story told on two graphs. The lower graph is the function you are integrating, f(x); the upper graph stays empty until the steps find the answer F(x). The last step of every indefinite integral checks the answer by differentiating it on the picture: the slope of F at any x equals the height of f there. Definite integrals shade the area, cut it into strips and compare it with F(b) − F(a). Press Next step (or Play). Under the controls each step is explained in detail, with links to the exact basics it uses; after reading a basic, press ← Back to return to the same step.
Guess a function whose derivative is the integrand, fix the constant factor, and add + C. Split sums and fractions into single powers first.
When a function and its derivative both appear, call the function t and the integral becomes a standard one. Identities lower powers and turn products into sums.
Complete the square to reach one of the six forms; when the top is linear, split it into (derivative of the bottom) + constant.
The dashed curves are the simple fractions; they add up to the integrand, and each one integrates to a log, a power or a tan⁻¹.
∫u v dx = u∫v dx − ∫(u′ ∫v dx) dx. Choose u so that it gets simpler when differentiated (ILATE), and watch for eˣ[f(x) + f′(x)].
The shaded area under f between the limits equals the rise of any antiderivative F between them. With substitution, change the limits too.
Split at sign changes, use even and odd symmetry, and the mirror property ∫f(x) dx = ∫f(a + b − x) dx.
Substitutions, partial fractions, parts twice, a clever split of √cot x + √tan x, and two definite integrals that need the properties.
A car’s odometer, a filling water tank, the work done stretching a spring, the average temperature of a day, the glass in an arched window and a workshop’s total cost: each one adds up a rate or a height, strip by strip. That total is an integral.
A car speeds up, cruises, then brakes to a stop. In each tiny time step dt it covers v · dt metres: a thin strip under the speed graph. Adding the strips gives the distance, ∫₀ᵗ v dt. The odometer reading is the green area.
Water flows in at r(t) = 20 + 10 sin(πt/6) litres per minute. The volume V(t) is an antiderivative of r, and the starting amount fixes the constant: V(0) = 50 gives V(t) = 50 + 20t + (60/π)(1 − cos(πt/6)). This is Example 4 with water.
A spring pulls back with force F = kx newtons when stretched x metres (here k = 200 N/m). Stretching it a further dx takes about F · dx joules of work, so the total work is ∫₀ˣ kx dx = ½kx², the area of the green triangle.
A day’s temperature T(t) rises and falls. Its average from midnight to hour b is (1/b)∫₀ᵇ T dt: the height of the rectangle whose area equals the area under the curve. The blue rectangle and the orange area always match.
The arch is the parabola y = h(1 − x²/a²) from x = −a to a. It is an even function, so the area is 2∫₀ᵃ h(1 − x²/a²) dx = (4/3)ah: exactly two-thirds of the rectangle around it, a fact Archimedes found over 2000 years ago.
A workshop’s marginal cost is MC(x) = 30 − 0.4x + 0.006x² rupees per item: the cost of one more. The total cost is the fixed cost (₹500 for the machine) plus the area under the MC curve: C(x) = 500 + ∫₀ˣ MC dx. This runs Chapter 6’s marginal cost backwards.
Decide first, then tap each card.
All 238 questions that ask you to find, evaluate or prove something have an animated walkthrough here; the 23 multiple-choice questions are in the Lesson’s self-check. Pick a question from the chips above each player. Press Play to watch it solved, or step through with Next. Each step has a short "why" under it and links to the basics it needs. Every indefinite answer ends with the slope check (the slope of F equals the height of f), and every definite answer ends with the area measured strip by strip.
Guess a function whose derivative is the integrand and fix the constant factor. Split sums and fractions into single powers first. (Q21–Q22 are in the quiz.)
Look for a function whose derivative is also a factor, call it t, and rewrite dx. (Q38–Q39 are in the quiz.)
Lower the powers with double angles, turn products into sums, and use 1 + tan²x = sec²x. (Q23–Q24 are in the quiz.)
The six standard forms, reached by completing the square; a linear top splits into (derivative of the inside) + constant. (Q24–Q25 are in the quiz.)
Divide if the top is not of lower degree; then one fraction per linear factor, two for a repeated factor, and a linear top over an irreducible quadratic. (Q22–Q23 are in the quiz.)
u from earlier in ILATE, v easy to integrate; parts twice when the integral comes back; and the eˣ[f + f′] shortcut. (Q23–Q24 are in the quiz.)
Complete the square, then use √(x² ± a²) or √(a² − x²). (Q10–Q11 are in the quiz.)
Find any antiderivative, evaluate it at the two limits, subtract. The top graph shows the rise of F; the bottom one the area. (Q21–Q22 are in the quiz.)
When you change the variable, change the limits with it and never go back. (Q9–Q10 are in the quiz.)
Split at sign changes, use odd and even symmetry, and add an integral to its mirror image. The dashed curve is the mirror. (Q20–Q21 are in the quiz.)
Q1–Q23 are indefinite integrals; Q24–Q31 definite integrals and Q32–Q37 proofs are in the second player. (Q38–Q40 are in the quiz.)
Integration runs differentiation backwards, and then measures area. To do it you need a handful of older ideas, each explained here from its roots: the derivative table read backwards, powers and roots, the trig identities, logs and e, the chain rule (which becomes substitution), completing the square, splitting fractions, area as a sum of thin strips, and symmetry. Every example and exercise links here exactly where you need it, and a button takes you straight back to the problem.
Integrating goes the other way from differentiating: given f, find an F whose slope at every x is f(x).
Differentiating turns a function F into its slope function F′ = f. So every derivative you know gives an integral for free.
Shifting a graph up or down by a constant C changes none of its slopes. Nothing in f can tell us which shift we started from, so the answer is the whole family F + C.
So ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C for every n ≠ −1.
∫cos = sin, ∫sin = −cos
∫sec²x = tan x
∫sec x tan x = sec x
∫cosec²x = −cot x
∫cosec x cot x = −cosec x
∫dx/√(1 − x²) = sin⁻¹x and ∫dx/(1 + x²) = tan⁻¹x.
∫eˣ dx = eˣ, ∫aˣ dx = aˣ/log a, ∫dx/x = log|x|.
Start from the product rule and integrate both sides:
That is integration by parts. The product rule also gives (eˣf)′ = eˣ(f + f′), the shortcut ∫ eˣ[f + f′] dx = eˣf + C.
The derivative rules themselves are proved on the Chapter 5 page.
The power rule only works on single powers. So first rewrite every root and fraction as a power of x.
Do this before you integrate anything.
These are the index laws. (x1/2)² = x¹, so x1/2 must be √x.
Why not n = −1? We would divide by 0. The antiderivative of x⁻¹ = 1/x is a new function, log x (Step 4).
(x³ − 1)/x² is not a power, but x³/x² − 1/x² = x − x⁻² is two powers. (√x − 1/√x)² expands to x − 2 + 1/x. Always rewrite before integrating.
On the unit circle the point at angle θ is (cos θ, sin θ). Pythagoras and the addition formula give everything else.
The first is Pythagoras on the unit circle. Divide it by cos²θ to get the second.
Put A = B in the addition formula cos(A + B) = cos A cos B − sin A sin B.
Rearranged, these lower the power. Adding and subtracting addition formulas turns products into sums.
sin²x = (1 − cos 2x)/2
cos²x = (1 + cos 2x)/2
sin³x = (3 sin x − sin 3x)/4
cos³x = (3 cos x + cos 3x)/4
= sin(A + B) + sin(A − B). Also
2 cos A cos B = cos(A + B) + cos(A − B)
2 sin A sin B = cos(A − B) − cos(A + B)
and cot²x = cosec²x − 1: both sec² and cosec² are on the table.
eˣ is the one exponential that is its own derivative, and log x (base e) undoes it.
Because the slope of log x is 1/x, the area under y = 1/t from 1 to x is log x.
The |x| covers negative x too, where log(−x) has the same derivative 1/x.
Use them to combine answers such as log|x + 1| − log|x + 2| = log|(x + 1)/(x + 2)|.
Whenever the top is the derivative of the bottom, ∫g′(x)/g(x) dx = log|g(x)| + C. Examples: ∫tan x dx = −log|cos x|, ∫2x/(1 + x²) dx = log(1 + x²).
When a function g(x) and its derivative g′(x) both appear, rename g(x) as t. This is the chain rule read backwards.
The first line is the chain rule. Read backwards, it gives the second.
This is just the derivative dt/dx = g′(x) with both sides multiplied by dx. The integral turns into ∫ f(t) dt.
The strips of width dx on the x-axis become strips of width dt = g′(x) dx on the t-axis: stretching the axis stretches the strips.
If the inside is ax + b, then dt = a dx and ∫f(ax + b) dx = (1/a)F(ax + b). For example ∫cos 3x dx = ⅓ sin 3x and ∫e^{2x+3} dx = ½e^{2x+3}.
Add and take away the right number, and any quadratic becomes a square plus a leftover constant.
Because (x + p)² = x² + 2px + p², x² + bx is (x + b/2)² with (b/2)² too many. With a leading coefficient a, take a out first.
After this shift, plus a square gives tan⁻¹ (or log|t + √(t² + k²)| under a root); minus a square gives a log (or sin⁻¹ when the t² has a minus sign).
Partial fractions undo the adding of fractions: they split one hard fraction into simple ones.
Left to right is the usual common denominator. Partial fractions go right to left.
If the top’s degree is not lower than the bottom’s, divide first, just like this.
One fraction per factor, two for a repeated factor (as here), and a linear top (Bx + C) over a quadratic factor that cannot be factorised.
Finding A, B: multiply out. The result is an identity, true for every x, so put in convenient x values (the cover-up trick) or compare coefficients.
For A in 1/((x − 1)(x + 2)) = A/(x − 1) + B/(x + 2): cover the factor (x − 1) and put x = 1 in what is left: A = 1/(1 + 2) = ⅓. Cover (x + 2) and put x = −2: B = 1/(−2 − 1) = −⅓.
A curved region has no simple area formula. So cut it into n thin strips and add them up.
A rectangle’s area is width × height. Each strip is nearly a rectangle of this width and height f(x) somewhere in the strip.
As n grows, the total settles on one number: the definite integral. Strips below the axis have negative height, so they count as negative area.
This holds for any antiderivative F. Why? The area from a to x, A(x), grows at the rate f(x): a thin extra strip of width h adds about f(x) · h. So A is an antiderivative.
A reflection does not change area. That one fact gives three shortcuts.
The graph is its own reflection in the y-axis (x², cos x). The area from −a to 0 equals the area from 0 to a.
Turn the graph half a turn about the origin and it lands on itself (x³, sin x). The two halves cancel.
The graph of f(a + b − x) is the reflection of f(x) in the middle line x = (a + b)/2, and a reflection does not change area.
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